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Class -XII PHYSICS
SQP Marking Scheme 2019-20
Section – A
1. a , ϕ= (for one face) 1
2. b , Conductor 1 3. a , 1Ω. 1
4. c ,12.0kJ 1 5. a , speed 1 6. d, virtual and inverted 1 7. a, straight line 1 8. d, 60 O 1 9. b, work function 1 10. b, third orbit 1 11. 45 O or vertical 1 12. 2 H 1 13. double 1 14. 1.227 Ao 1 15. 60° 1 16. Difference in initial mass energy and energy associated with mass of products
Or Total Kinetic energy gained in the process
1
17. Increases 1 18. No/8 1 19. 0.79 eV 1 20. Diodes with band gap energy in the visible spectrum range can function as LED 1
2
OR Any one use
Section – B
21. When electric field E is applied on conductor force acting on free electrons = -e
m = -e
=
Average thermal velocity of electron in conductor is zero (ut)av= 0 Average velocity of electron in conductors in τ (relaxation time) = vd (drift velocity) vd = (ut)av + a τ vd = 0 +
=
1 1
22.
C2 and C3 are in series
= + = 1
= 1μf
& C4 are in ∥
C” = 1 + 1 = 2μf C” & c5 are in series
= + ⟹ = 1μf
& c1 are in ∥
Ceq = 1 + 1 = 2μf Energy stored
U = cv2 = ×2×10-6×62
= 36×10-6J
1 1
3
23. Gain in KE of particle = Qv
= KP = qpVp ----------(i)Vp = V∝ =V
= K∝ = q∝V∝ -----------(ii)
(ii)/(i)
= =
= = =
v∝ : vp = 1:
1 1
24. “The angle of incidence at which the reflected light is completely plane
polarized, is called as Brewster’s angle (iB)
At i = iB, reflected beam 1 to refracted beam
∴ iB + r = 90 ⟹ r = 90-iB Using snell’s law = μ
= μ ⟹ = μ
μ = tan
1 1
25. wave function
ω = 2.14eV (a) Threshold frequency ω = hν0
ν0 = =
1
4
= 5.17× Hz (b) As kmax = eV0 = 0.6eV
Energy of photon E = kmax + ω = 0.6eV + 2.14eV = 2.74eV
Wave length of photon λ = =
= 4530Å
1
26.
centripetal force = electrostatic attraction
=
= --------(i)
as = n .
= put in (i)
m . =
=
OR
Energy of electron in n = 2 is -3.4eV
∴ energy in ground state = -13.6eV
kE = -TE = +13.6eV
1 1 1
En = ⇒ -3.4eV = ⇒
energy in ground state x = - 13.6eV.
5
PE = 2TE = -2×13.6eV = -27.2eV
1
27.
P-type semiconductor n-type semiconductor 1. Density of holes >> density of
electron
2. Formed by doping trivalent impurity
1. density of electron>>density of holes
2. formed by doping pentavalent impurity
Energy band diagram for p-type
Energy band diagram of n-type semiconductor
OR
Energy of photon E = = eV =2.06eV
As E
6
decreased and hence potential gradient ‘k’ will also be decreased. Thus the null point or balance point will shift to right (towards, B) side.
29.
According to Biot-Savart’s law, magnetic field due to a current element is given by
= where r =
∴ dB =
And direction of is ⊥ to the plane containing I and .
Resolving along the x – axis and y – axis.
dBx = dB sin
dBy = dB cos
taking the contribution of whole current loop we get
Bx = ∮dBx = ∮dBsin = .
Bx = ∮dl =
And By = ∮dBy = ∮dBcos = 0
∴ BP = = Bx =
∴ = = )
For centre x = 0
∴ = = in the direction of
1 1/2 1/2 1
7
30.
resonant frequency for LCR circuit is given by =
=
= 17.69Hz
Or = = 111rad/s.
quality factor of resonance
Q = = =
∴ Q = = 45.0
To improve sharpness of resonance circuit by a factor 2, without reducing ; reduce R to half of its value i.e. R = 3.7Ω
1 1 1
31.
Two conditions for T IR –
(a) Light must travel from denser to rarer medium (b) i>ic
Sin ic =
∴ (ic)Red = Sin-1 = 46°
(ic) Green= Sin-1 = 44.8°
(ic)Blue = Sin-1 = 43°
Angle of incidence at face AC is 45° which is more than the critical angle for Blue
and Green colours therefore they will show TIR but Red colour will refract to other medium.
1 1 1
32. Resolving power (R.P) of an astronomical telescope is its ability to form separate
images of two neighboring
astronomical objects like stars etc.
R.P. = = where D is diameter of objective lens and is wave length
1
8
of light used.
D = 100inch = 2.54×100cm = 254cm
= 2.54m
= 6000Å
Limit of resolution =
=
= 2.9×
OR Basic assumptions in derivation of Lens-maker’s formula: (i) Aperture of lens should be small (ii) Lenses should be thin (iii) Object should be point sized and placed on principal axis.
Suppose we have a thin lens of material of refractive index n2, placed in a medium of refractive index n1, let o be a point object placed on principle axis then for refraction at surface ABC we get image at I1 , ∴ - = ---------(1)
But the refracted ray before goes to meet at I1 falls on surface ADC and refracts at I2
1 1 1 1
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finally; hence I1 works as a virtual object 2nd refracting surface ∴ - = --------- (2)
Equation (1) + (2) - = (
∴ = ( --------(3)
If u = ∞, ѵ = f = ( --------(4)
Which is lens maker’s formula.
1
33. → + + Q
Q = [MU – MPa – MH] c2 = [ 238.05079 – 237.05121 – 1.00783] u × c2 = - 0.00825u × 931.5
= - 7.68MeV Q
10
Section – D 35. (a)
According the Gauss’s law –
∮N, d s
= {q}
∫ E
+ ∫ E
+ ∫E
= (λL)
∫Eds1Cos0 + ∫Eds2Cos90o+ ∫Eds3Cos90o =
E∫ds1 =
E × 2πrL =
E =
E
=
1 1 1
35. (b)
∵ Ex = ∝ x = 400x Ey = Ez = 0
Hence flux will exist only on left and right faces of cube as Ex≠ 0 ∵ . a2 ) + . a2 = {qin} =
- . a2 ) + a2 = = -(400a)a2 + a2 (400 × 2a) = -400a3 + 800a3 = 400a3 = 400 × (.1)3 = 0.4 Nm2c-1
1
11
(a) (b)
∵ = {qin}
∴ qin =
= 8.85 × 10-12 × 0.4 = 3.540×10-12c
OR
Definition of electrostatic potential – SI unit J/c of Volt. Deduction of expression of electrostatic potential energy of given system of charges –
U =
1 1 2 1 1
36.
For forward motion from x = 0 to x = 2b. The flux ϕB linked with circuit SPQR is
12
ϕB = Blx 0≤x
13
OR Working principle of cyclotron Diagram Working of cyclotron with explanation Any two appliations
1 1 2 1
37.
Deduction of mirror formula + =
For a convex mirror f is always +ve.
∴ f > c
Object is always placed in front of mirror hence u < 0 (for real object)
+ =
⇒ = -
As u < 0 u –ve hence
> 0
⇒ v> 0 i.e. +ve for all values of u. Image will be formed behind the mirror and it will be virtual for all values of u.
1 2 1 1
37. (a)
OR Ray Diagram : (with proper labeling)
1
14
Magnifying power m =
m =
1
37. (b)
∵ m = mo me = -30 (virtual, inverted)
∵fo = 1.25cm
fe = 5.0cm
Let us setup a compound microscope such that the final image be formed at D, then
me = 1 + = 1 + = 6
and position of object for this image formation can be calculated –
- =
- =
- = + =
= = - 4.17cm.
∵ m = mo × me
∴ mo = = = -5
∴ V = -5uo
- =
- =
1 1
15
=
uo = -1.5cm ⟹vO = 7.5cm
Tube length = Vo +|ue| = 7.5cm + 4.17cm
L = 11.67cm
Object should be placed at 1.5cm distance from the objective lens.
1
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